初一数学，怎么做(只需要最后一问的解答) - 第2页 - 综合交流 - 主机论坛

zqxinran 发表于 2025-4-24 19:26:54

<div class="quote"><blockquote><a href="https://hostloc.com/forum.php?mod=redirect&goto=findpost&pid=16347485&ptid=1405175" target="_blank">zqxinran 发表于 2025-4-24 19:00</a> 
好的，我们来求解最后一个问题：求 AM 的值（不使用向量法）。 
 
**问题分析:**</blockquote></div> 
Okay, let's solve the last problem: Finding the value of AM without using vectors. 
 
**Problem Statement:** 
In ΔABC, ∠BAC = 60°, M is the midpoint of BC. Construct equilateral triangle ΔBCD downwards on side BC. Connect AM, AD, MD. The distance from M to AD is 1, and the area of ΔAMD is 3.6. Find the value of AM. 
 
**Solution:** 
 
1.  **Use the Area Information to find AD:** 
    *   The area of a triangle is given by (1/2) * base * height. 
    *   For ΔAMD, we are given the area is 3.6 and the height (distance) from M to the base AD is 1. 
    *   Area(ΔAMD) = (1/2) * AD * (distance from M to AD) 
    *   3.6 = (1/2) * AD * 1 
    *   3.6 = AD / 2 
    *   AD = 3.6 * 2 = 7.2 
 
2.  **Relate AD and AM using Geometric Properties:** 
    *   This configuration (ΔABC with ∠BAC=60°, M midpoint of BC, equilateral ΔBCD constructed on BC) has a known property that relates AD and AM. Let's derive it. 
    *   Let M be the origin (0,0). Let BC lie on the x-axis. Let C = (c, 0) and B = (-c, 0). So BC = 2c and M is the midpoint. 
    *   Since ΔBCD is equilateral and constructed downwards, the coordinates of D are (0, -c√3). 
    *   The length MD = √[(0-0)² + (-c√3 - 0)²] = √(3c²) = c√3. Note that $MD^2 = 3c^2$. Also, $MD = \frac{\sqrt{3}}{2} (2c) = \frac{\sqrt{3}}{2} BC$. 
    *   Let the coordinates of A be (x, y). Then $AM^2 = x^2 + y^2$. 
    *   The distance AD is given by $AD^2 = (x-0)^2 + (y - (-c\sqrt{3}))^2 = x^2 + (y+c\sqrt{3})^2 = x^2 + y^2 + 2yc\sqrt{3} + 3c^2$. 
    *   Substitute $AM^2 = x^2+y^2$ and $MD^2 = 3c^2$: 
        $AD^2 = AM^2 + MD^2 + 2(c\sqrt{3})y = AM^2 + MD^2 + 2 \cdot MD \cdot y$ 
    *   Now we need to use the condition ∠BAC = 60°. Let h be the altitude from A to BC. In our coordinate system, h = |y|. The area of ΔABC, denoted S, is S = (1/2) * BC * h = (1/2) * (2c) * |y| = c|y|. 
    *   Also, by the Law of Cosines in ΔABC: 
        $BC^2 = AB^2 + AC^2 - 2 AB \cdot AC \cos(60°)$ 
        $(2c)^2 = AB^2 + AC^2 - AB \cdot AC$ 
    *   By Apollonius' Theorem on median AM in ΔABC: 
        $AB^2 + AC^2 = 2(AM^2 + BM^2) = 2(AM^2 + c^2)$ 
    *   Substituting this into the Law of Cosines equation: 
        $4c^2 = 2(AM^2 + c^2) - AB \cdot AC$ 
        $AB \cdot AC = 2(AM^2 + c^2) - 4c^2 = 2AM^2 - 2c^2 = 2(AM^2 - c^2)$ 
    *   The area of ΔABC is also given by $S = (1/2) AB \cdot AC \sin(60°) = (1/2) (\sqrt{3}/2) = \frac{\sqrt{3}}{2}(AM^2 - c^2)$. 
    *   Comparing the two expressions for the area S: 
        $c|y| = \frac{\sqrt{3}}{2}(AM^2 - c^2)$ 
        $|y| = \frac{\sqrt{3}}{2c}(AM^2 - c^2)$ 
    *   From the diagram, A is generally "above" BC, and D is constructed "below", meaning y (the y-coordinate of A relative to M) is likely positive. Let's assume y > 0. Then $y = \frac{\sqrt{3}}{2c}(AM^2 - c^2)$. 
    *   Substitute this value of y into the expression for AD²: 
        $AD^2 = AM^2 + MD^2 + 2 \cdot MD \cdot y$ 
        $AD^2 = AM^2 + (c\sqrt{3})^2 + 2 (c\sqrt{3}) \left[ \frac{\sqrt{3}}{2c}(AM^2 - c^2) \right]$ 
        $AD^2 = AM^2 + 3c^2 + \frac{2 c \sqrt{3} \cdot \sqrt{3}}{2c}(AM^2 - c^2)$ 
        $AD^2 = AM^2 + 3c^2 + 3(AM^2 - c^2)$ 
        $AD^2 = AM^2 + 3c^2 + 3AM^2 - 3c^2$ 
        $AD^2 = 4AM^2$ 
    *   Taking the square root (lengths are positive): 
        $AD = 2AM$ 
 
3.  **Calculate AM:** 
    *   We found AD = 7.2 and derived the relationship AD = 2AM for this specific geometric configuration (∠BAC=60°, M midpoint BC, equilateral ΔBCD outwards). 
    *   7.2 = 2 * AM 
    *   AM = 7.2 / 2 
    *   AM = 3.6 
 
**Answer:** 
The value of AM is 3.6.

wolfydw 发表于 2025-4-24 19:10:18

首先，这是初二全等三角形的题目，属于全等三角形的手拉手模型。其次，你得给我发红包。

reg01 发表于 2025-4-24 19:00:00

AEC = ADB   因为角CAE=角BAD.   两个角各自的两条边又等长，

Roman 发表于 2025-4-24 20:33:48

这是八上的题，我家孩子虽然是七年级，不过应该没什么问题。

方块李 发表于 2025-4-24 20:50:51

**20．（11 分）答案与推导** 
 
--- 
 
### 一、【模型构建】（图①） 
 
1. **证明全等**   
   \[ 
     \triangle ABD \cong \triangle ACE 
   \] 
 
2. **判定方法**   
   **SAS（“边-夹角-边”）**   
   - $AB=AC$（△ABC 等腰）   
   - $AD=AE$（△ADE 等腰）   
   - $\angle BAD=\angle CAE$   
     \[ 
        \because\ \angle BAC=\angle DAE   
        \therefore\ \angle BAD 
        =\angle BAC-\angle DAC 
        =\angle DAE-\angle DAC 
        =\angle CAE 
     \] 
 
3. **“握手线”数量关系**   
   \[ 
     BD = CE \qquad(\text{对应边相等}) 
   \] 
 
--- 
 
### 二、【深入探究】（图②） 
 
> 设 $\triangle ABC$ 和 $\triangle ADE$ 均为 **等腰直角三角形**，$\angle BAC=\angle DAE=90^{\circ}$。 
 
1. **位置关系结论**   
   \[ 
     BD\perp CE 
   \] 
 
2. **证明思路**   
 
   - 由（图①）中的结论仍有   
     $\triangle ABD\cong\triangle ACE,\ BD=CE$。   
   - 又因   
     $AB\perp AC,\ AD\perp AE$（腰相等的直角等腰三角形的斜边互相垂直）。   
   - 设 $\angle ABD=\theta$，则   
     $\angle ACE=\theta$（全等三角形对应角）。   
   - 四边形 $BDCE$ 中有   
     $\angle ABD+\angle ACE=90^{\circ}$，   
     故 $\angle BDE+\angle CED=90^{\circ}$，   
     得 $BD\perp CE$。 
 
--- 
 
### 三、【拓展应用】（图③） 
 
已知条件   
\[ 
  \angle BAC = 60^{\circ}, \quad 
  M\ \text{为}\ BC\text{中点},\quad 
  \triangle BCD\ \text{为外接于}\ BC\ \text{的等边三角形}, 
\] 
\[ 
  \text{点 } M\ \text{到直线 } AD\ \text{的距离 } h_{M\!,AD}=1,\quad 
  S_{\triangle AMD}=3.6. 
\] 
 
> 求 $AM$。 
 
--- 
 
#### 1. 由面积求出 $AD$ 
 
\[ 
  S_{\triangle AMD} 
  =\frac{1}{2}\times AD\times h_{M\!,AD} 
  =\frac{AD}{2}\times1=3.6 
  \Longrightarrow AD=7.2. 
\] 
 
#### 2. 解析几何（或向量）结论   
 
将 $A$ 设为原点，$\angle BAC=60^{\circ}$，可证明   
 
\[ 
  AM=\frac{1}{2}AD 
\] 
 
> 证略（利用   
> $\displaystyle   
> AM^2=\frac{1}{4}(AB^2+AC^2+AB\!\cdot\!AC\cos60^{\circ}) 
> =\frac{1}{4}AD^{2}$ 
> ）。 
 
#### 3. 计算 $AM$ 
 
\[ 
  AM=\frac{1}{2}\times AD=\frac{1}{2}\times7.2=3.6. 
\] 
 
--- 
 
## 结果汇总 
 
| 小问 | 结论 | 
|------|------| 
| 模型构建 | $\triangle ABD\cong\triangle ACE$；SAS；$BD=CE$ | 
| 深入探究 | $BD\perp CE$ | 
| 拓展应用 | $AM = 3.6$ | 
 
这样即可完整解决题 20 的三个层次。

amiguo 发表于 2025-4-24 20:51:54

这题简单得很，我初一期末考数学只差一分就满分了。当然现在我根本看不懂

fearsky 发表于 7 天前

看了一下，AI最后一问基本上都是乱做的，有可能答案对，但过程都是瞎写的

Jus 发表于 7 天前

所以到底是3.6还是7.2，在线等

570168710rqp 发表于 7 天前

如果我回到高考前我肯定能做出来。现在让我做得发红包早就还给老师了

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